What is the derivative of #log(sin^2(x))#?

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Answer 1 · 2015-08-31T18:53:27

If you don't remember the derivative of #log_10(u)#, you can derive it like so:

#y = log(sin^2x)#
#10^y = sin^2x#

#d/(dx)[10^y] = 10^yln10((dy)/(dx))#

(It is an extension of #d/(dx)[a^x] = a^xlna#.)

Therefore, you get:

#10^yln10((dy)/(dx)) = 2sinxcosx#

#color(blue)((dy)/(dx)) = (2sinxcosx)/(10^yln10)#

#= (2sinxcosx)/(sin^2x*ln10)#

#color(blue)(= (2cotx)/(ln10))#